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(B) The magnetic field at the centre of a circular coil of radius $R$ carrying current $I$ is given by $B = \frac{\mu_0 I}{2 R}$.For coil $P$ with cur

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$\frac{3 \mu_0 I}{2 R}$

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(B) The magnetic field at the centre of a circular coil of radius $R$ carrying current $I$ is given by $B = \frac{\mu_0 I}{2 R}$.For coil $P$ with current $I$,the magnetic field is $\vec{B_P} = \frac{\mu_0 I}{2 R}$.For coil $Q$ with current $\sqrt{8} I$,the magnetic field is $\vec{B_Q} = \frac{\mu_0 \sqrt{8} I}{2 R}$.Since the coils are in perpendicular planes,their magnetic fields $\vec{B_P}$ and $\vec{B_Q}$ are perpendicular to each other.The magnitude of the net magnetic field is $B_{	ext{net}} = \sqrt{B_P^2 + B_Q^2}$.Substituting the values: $B_{	ext{net}} = \sqrt{\left(\frac{\mu_0 I}{2 R}ight)^2 + \left(\frac{\mu_0 \sqrt{8} I}{2 R}ight)^2}$.$B_{	ext{net}} = \frac{\mu_0 I}{2 R} \sqrt{1^2 + (\sqrt{8})^2} = \frac{\mu_0 I}{2 R} \sqrt{1 + 8} = \frac{\mu_0 I}{2 R} \sqrt{9} = \frac{3 \mu_0 I}{2 R}$.

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